If $x=\log \frac{3}{5}+\log \frac{5}{2}+\log \frac{7}{27}$ then

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

x=\log \frac{3}{5}+\log \frac{5}{2}+\log \frac{7}{27}

then

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Correct option: D

x \geq \frac{3}{(3)^{1 / 3}}

Option D (

x \geq \frac{3}{(3)^{1 / 3}}

) is correct because by applying AM >= GM to the given logarithmic expression, we arrive at this inequality.

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}